Question

Solve the three algebraic expressions:

& 5x-6y+4z=15 \\\ & 7x+4y-3z=19 \\\ & 2x+y+6z=46 \\\ \end{aligned}$$

Answer 1

Hint: Consider any 2 of the expressions and convert the 3 variables to 2 variables. Similarly find 2 equations and convert 3 variables to 2 variables.We get 2 equations having 2 variables and Solve them,we get values of two variables and substitute in any equation to get the value of another variable.Hence desired values of x ,y and z can be found.

Complete step-by-step answer:
We have been given three algebraic expressions. Now let us pick any two pairs of equations of the system.

& 5x-6y+4z=15-(1) \\\ & 7x+4y-3z=19-(2) \\\ & 2x+y+6z=46-(3) \\\ \end{aligned}$$ Let us take equation (2) and equation (3). Then use addition or subtraction to eliminate the same variable from both pairs of equations. $$\begin{aligned} & 7x+4y-3z=19-(2)\times 2 \\\ & 2x+y+6z=46-(3)\times 7 \\\ \end{aligned}$$ Let us subtract both these equations and we will get an equation with 2 variables. Multiply 2 in equation (2) and 7 in equation (3). 14x + 8y - 6z = 38 14x + 7y + 42z = 322 y – 48z = -284 $$\therefore y-48z=-284-(4)$$ Now let us take equation (1) and equation (3), multiply equation (3) by 5 and subtract both the equations. $$\begin{aligned} & 5x-6y+4z=15-(1)\times 2 \\\ & 2x+y+6z=46-(3)\times 5 \\\ \end{aligned}$$ Thus we get new equations as, 10x – 12y + 8z = 30 10x + 5y + 30z = 230 \- 17y -22z = -200 $$\Rightarrow 17y+22z=200-(5)$$ Now let us solve equation (4) and (5) to get the value of z. $$\begin{aligned} & y-48z=-284-(4) \\\ & 17y+22z=200-(5) \\\ \end{aligned}$$ Multiply equation (4) $$\times 17$$. Equation (4) becomes, $$17y-816z=-4828$$. 17y – 816z = -4828 17y + 22z = 200 0 – 838z = -5028 $$\begin{aligned} & \therefore -838z=-5028 \\\ & \therefore z=\dfrac{5028}{838}=6 \\\ \end{aligned}$$ Hence we got the value of z = 6, substitute this value in equation (4). $$\begin{aligned} & y=48z-284=48\times 6-284=4 \\\ & \therefore y=4 \\\ \end{aligned}$$ Substitute y = 4 and z = 6 in equation (1). $$\begin{aligned} & 5x=15+6y-4z \\\ & 5x=15+\left( 6\times 4 \right)-\left( 4\times 6 \right) \\\ & \therefore x=\dfrac{15}{5}=3 \\\ \end{aligned}$$ Thus we got the value as, x = 3, y = 4 and z = 6. Hence we solved the algebraic expressions. Note: An algebraic expression can have any number of variables. Here we were given 3 variables x, y and z. Thus we required 3 equations to solve and find the values. For an expression of 2 variables only 2 equations are required.